Question: Let $a(x)=5x^3+2x^2+x+2$, and $b(x)=x^3+x+1$. When dividing $a$ by $b$, we can find the unique quotient polynomial $q$ and remainder polynomial $r$ that satisfy the following equation: $\dfrac{a(x)}{b(x)}=q(x) + \dfrac{r(x)}{b(x)}$, where the degree of $r(x)$ is less than the degree of $b(x)$. What is the quotient, $q(x)$ ? $ q(x)=$ What is the remainder, $r(x)$ ? $r(x)=$
Note that $a(x)$ has a higher degree than $b(x)$. This allows us to find a non-zero quotient polynomial, $q(x)$. [Why is this important?] Let's use long division with polynomials in order to find the quotient, $q(x)$ and remainder, $r(x)$ of $\ \dfrac{a(x)}{b(x)}=\dfrac{5x^3+2x^2+x+2}{x^3+x+1}$ : We divide ${x^3}$ into ${5x^3}$ to get ${5}$ : $ \hphantom{1567|14444} {5}\\ {{{x^3}+x+1}}|\overline{{5x^3}+2x^2+x+2}\\ \hphantom{37..........|}\llap{-}\underline{(5x^3+0x^2+5x+5)}\\ \hphantom{37|3....999......}{+2x^2-4x -3}\\ $ [What did we do here?] The process stops here because $x^3+x+1$ is a polynomial of the third degree and $2x^2-4x-3$ is a polynomial of the second degree. So it follows that ${r(x)}={2x^2-4x-3}$, ${q(x)}={5}$, and $ \dfrac{5x^3+2x^2+x+2}{x^3+x+1}={5}+\dfrac{{2x^2-4x-3}}{x^3+x+1}$ To conclude, $q(x)=5$ $r(x)=2x^2-4x-3$